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Unformatted text preview: lee (cyl327) HW04 gentle (56245) 1 This printout should have 33 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A point on the outer rim of a tire on a moving vehicle exhibits uniform circular motion with a frequency of 28 Hz. The diameter of the tire is 35 . 2 cm. How fast is the car moving? Correct answer: 111 . 469 km / hr. Explanation: You first need to find the length of one cycle, which is the circumference of the tire. Thus = d Thus the velocity is defined by v = f Conversion for v: 1 s cm 1 1 m 100 cm 1 km 1000 m 3600 s 1 hr = km / hr 002 (part 1 of 2) 10.0 points Young David, who slew Goliath, experi mented with slings before tackling the gi ant. He found that with a sling of length . 253 m, he could revolve the sling at the rate of 7 . 9 rev / s. If he increased the length to 0 . 637 m, he could revolve the sling only 4 . 7 rev / s. a) What is the larger of the two linear speeds? Correct answer: 18 . 8112 m / s. Explanation: Basic Concept: The angular velocity is defined as v r , in units of rad s . Solution: v = r , so v 1 = r 1 1 = (0 . 253 m) (7 . 9 rev / s) parenleftbigg 2 rad rev parenrightbigg = 12 . 5582 m / s . and v 2 = r 2 2 = (0 . 637 m) (4 . 7 rev / s) parenleftbigg 2 rad rev parenrightbigg = 18 . 8112 m / s . Therefore the greater linear speed is 18 . 8112 m / s. 003 (part 2 of 2) 10.0 points b) Using the sling length 0 . 637 m, what is the centripetal acceleration at 4 . 7 rev / s? Correct answer: 555 . 514 m / s 2 . Explanation: The centripetal acceleration is given by a 2 = v 2 2 r 2 = r 2 2 2 = (0 . 637 m) (4 . 7 rev / s) 2 parenleftbigg 2 rad 1 rev parenrightbigg 2 = 555 . 514 m / s 2 . 004 (part 1 of 2) 10.0 points The orbit of a Moon about its planet is ap proximately circular, with a mean radius of 4 . 08 10 8 m. It takes 39 . 3 days for the Moon to complete one revolution about the planet. Find the mean orbital speed of the Moon. Correct answer: 754 . 977 m / s. Explanation: Dividing the length C = 2 r of the trajectory of the Moon by the time T = 39 . 3 days = 3 . 39552 10 6 s lee (cyl327) HW04 gentle (56245) 2 of one revolution (in seconds!), we obtain that the mean orbital speed of the Moon is v = C T = 2 r T = 2 (4 . 08 10 8 m ) 3 . 39552 10 6 s = 754 . 977 m / s . 005 (part 2 of 2) 10.0 points Find the Moons centripetal acceleration. Correct answer: 0 . 00139704 m / s 2 . Explanation: Since the magnitude of the velocity is con stant, the tangential acceleration of the Moon is zero. The centripetal acceleration is a c = v 2 r = (754 . 977 m / s ) 2 4 . 08 10 8 m = 0 . 00139704 m / s 2 ....
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This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas at Austin.
 Fall '10
 MARKERT
 mechanics

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